I recently listened to @mrhonner’s episode of @myfavethm, in which he cited Varignon’s Theorem as his favourite.

### What’s Varignon’s Theorem when it’s at home?

It states that, if you draw any quadrilateral, then connect the midpoints of adjacent sides, you get a parallelogram. Don’t believe it? Try Mark’s nifty geometry tool and see for yourself.

That’s pretty neat, and Mr Honner gave a nice proof of it in the podcast - but it struck me as the kind of thing that a question, either at A-level or the top end of GCSE, might ask one to prove using vectors.

### So, I present to you: A Varignon Vector Masterclass.

Suppose I draw the quadrilateral OABC, and let $\vec{OA}=2\bb a$, $\vec{OB}=2\bb b$ and $\vec{OC}=2\bb c$. (I’ve picked double-vectors because I know I’ll be halving them shortly. While I like fractions, there’s no need to make typesetting harder than it needs to be.)

I’ll also make P, Q, R and S the midpoints of $\vec{OA}$, $\vec{AB}$, $\vec{BC}$ and $\vec{CO}$, respectively. The connections between midpoints are PQ, QR, RS and SP.

The vector $\vec{PQ}$ is $\bb a + (\bb b-\bb a) = \bb b$.

The vector $\vec{SR}$ is $\bb c + (\bb b- \bb c)=\bb b$, so one pair of opposite sides of PQRS are equal in length and parallel - making PQRS a parallelogram!

(If you’re not convinced this is enough, you can repeat the logic with PS and QR to show that both pairs of opposite sides are equal - but it’s not necessary.)

Here’s a diagram you can play with.

### Interestingly

(… or not, depending on your point of view), I didn’t make any assumptions about the positions of the points - the quadrilateral they define isn’t necessarily convex, it isn’t necessarily simple (it could cross itself) and it isn’t even necessarily planar - it would work with any four points in three dimensions.

I’m now getting worried that all the good theorems are taken, and if Evelyn and Kevin invite me on their podcast, I’ll have nothing to talk about!